Identify the substrate

The compound is chloromethyl methyl ether: $\mathrm{CH_3-O-CH_2-Cl}$

The carbon bearing chlorine is primary, but it is also adjacent to oxygen.

Examine Statement I: It undergoes $S_N1$ though it is a primary halide

Normally, primary halides do not undergo $S_N1$ because formation of a primary carbocation is unstable.

But here, loss of $\mathrm{Cl^-}$ gives a cation next to oxygen: $\mathrm{CH_3-O-CH_2^+}$

This cation is stabilized by resonance with the lone pair on oxygen: $\mathrm{CH_3-O-CH_2^+ \leftrightarrow CH_3-O^+=CH_2}$

Thus, the intermediate is not an ordinary unstable primary carbocation; it is resonance-stabilized oxonium/carboxonium-type cation.

Therefore, $S_N1$ is possible.

So, Statement I is correct.

Examine Statement II: It will not undergo $S_N2$ very easily though it is a primary halide

At first glance, a primary halide should favor $S_N2$.

However, in this molecule, the carbon attached to chlorine is bonded to an oxygen-containing group: $\mathrm{-OCH_3}$

Oxygen exerts a strong $-I$ effect, making the carbon electrophilic, but the adjacent oxygen also affects backside attack because the substrate is a type of alkoxy methyl halide, where the reaction pathway can be influenced by neighboring electronic effects and easier ionization.

More importantly, such compounds are known to undergo substitution not predominantly by a simple easy $S_N2$ route; due to stabilization of the cation by oxygen, $S_N1$ becomes unusually favorable, and $S_N2$ is not very easy compared with normal primary alkyl halides.

Hence, Statement II is also correct.

Conclusion

• Statement I: Correct
• Statement II: Correct

Therefore, the correct option is: $\boxed{\text{B}}$

Comparison with stored answer

Stored correct answer: B

My derived answer: B

So, they agree.